我们这里设置一个头结点,然后遍历两个链表,使用一个flag记录相加的结果和进位,如果两个链表没有走到最后或者进位不等于0,我们就继续遍历处理进位;如果当前的链表都遍历完成了,判断当前的进位是否>10,然后处理是否需要添加进位结点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* newhead = new ListNode(-1);
ListNode* pthread = newhead;
ListNode* cur1 = l1;
ListNode* cur2 = l2;
int flag = 0;
while(cur1 || cur2 || flag)
{
if(cur1)
{
flag += cur1->val;
cur1 = cur1->next;
}
if(cur2)
{
flag += cur2->val;
cur2 = cur2->next;
}
ListNode*node = new ListNode(flag % 10);
flag /= 10;
pthread->next = node;
pthread = node;
}
pthread = newhead ->next;
delete newhead;
return pthread;
}
};